Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

----- (-e)-------a-------(e)------

-----(-e)--------b-------(e)-----

Oct 22nd 2009, 10:47 PM

redsoxfan325

Quote:

Originally Posted by mtlchris

Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

----- (-e)-------a-------(e)------

-----(-e)--------b-------(e)-----

We may assume wlog that . Thus the open interval around is

The open interval around is

From this it's clear that and are disjoint, so .

Oct 22nd 2009, 10:51 PM

mtlchris

what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?

Oct 22nd 2009, 10:55 PM

redsoxfan325

Quote:

Originally Posted by mtlchris

what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?

It doesn't have to be; take any , and taking the radius to be , will guarantee .

Oct 22nd 2009, 11:01 PM

mtlchris

I don't really follow...how come we can assume that a < b ?

Oct 22nd 2009, 11:04 PM

redsoxfan325

Quote:

Originally Posted by mtlchris

I don't really follow...how come we can assume that a < b ?

One of them has to be greater than the other. If it's not , then it's . The reason we can assume wlog is that if we assumed , we would still get the same result.

Oct 22nd 2009, 11:10 PM

mtlchris

and we make this assumption so we can relate U and V?

Oct 22nd 2009, 11:11 PM

redsoxfan325

I made the assumption so I didn't have to deal with the absolute value sign.

Oct 22nd 2009, 11:32 PM

mtlchris

ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R

is this correct? and if so....how does he go from abs(c-a) < e to
abs(x-a) < e
and then to = 1/2 abs(a-b)

Oct 22nd 2009, 11:37 PM

redsoxfan325

Quote:

Originally Posted by mtlchris

ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R

is a set; is a number. How can those be equal?

Oct 22nd 2009, 11:42 PM

mtlchris

ok i should have written
{ c in R : abs(c-x) < e }

Oct 23rd 2009, 12:06 AM

redsoxfan325

Quote:

Originally Posted by mtlchris

ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R